Question

Use the differentiation rules developed in this section to find the derivatives of the functions in Exercises 35-64. Note that it may be necessary to do some preliminary algebra before differentiating.

$f\left(x\right)=\frac{x-1}{(x+1)(x+2)}$

Short answer

The derivative of the function is$f\text{'}\left(x\right)=\frac{-x^2+2x+5}{{(x^2+3x+2)}^2}$

Step-by-step solution

Step 1. Given Information

The given function is$f\left(x\right)=\frac{x-1}{(x+1)(x+2)}$.

Step 2. Simplify the function

Simplify the denominator of the function.

$\begin{array}{rcl}f\left(x\right)& =& \frac{x-1}{x(x+2)+1(x+2)}\\ & =& \frac{x-1}{x^2+2x+x+2}\\ & =& \frac{x-1}{x^2+3x+2}\end{array}$

Step 3. Find the derivative

Apply the quotient rule of derivative, $\left(\frac{f}{g}\right)\text{'}\left(x\right)=\frac{f\text{'}\left(x\right)g\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{(g{\left(x\right))}^2}$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{{\displaystyle \frac{d}{dx}}(x-1)·(x^2+3x+2)-(x-1){\displaystyle \frac{d}{dx}}(x^2+3x+2)}{{(x^2+3x+2)}^2}\\ & =& \frac{{\displaystyle \left(1\right)·(x^2+3x+2)-(x-1)(2x+3)}}{{(x^2+3x+2)}^2}\\ & =& \frac{x^2+3x+2-(2x^2+3x-2x-3)}{{(x^2+3x+2)}^2}\\ & =& \frac{-x^2+2x+5}{{(x^2+3x+2)}^2}\end{array}$